First Year 'A' Level Chemistry Page 1 of "Chapter 34 : The shapes of molecules and Ions"

Valence Shell Electron Pair Repulsion theory (VSEPR), A Y1/Y2 Blog for the week commencing 15th March 2020. This blog is about how Valence Shell Electron Pair Repulsion (VSEPR) theory determines the Shape of Molecules and of Ions. • The starting point for Shapes are Lewis Structures and Formal Charges, therefore please revise them

(cf. Blog dated 20th October 2019). A LITTLE BIT ABOUT GEOMETRY • Let me start by telling you that in Geometry one object (let’s say a miniscule point such as an electron)

cannot constitute a shape – and the same is true in VSEPR theory. However, once you have two objects (say two points or two bonded atoms), then these two objects now form a shape, and the shape that they form is a straight line because you can join the two points together by a straight line. (In VSEPR theory, a straight line is called a “linear” shape.) To make any other shape out of two points is NOT possible because to make a triangle you would need three points, and to make a square you would need four points, and to make a squiggly line you would need a HUGE number of points ...... and so on. OK, that is nice and easy, and it is also easy when there are only three objects/points because as we saw just now, three objects/points would give either a straight line or a triangle – but with four or more points, things can get a bit difficult.

• Now, I need to tell you something that I cannot prove because I am not sufficiently good at Euclidean

Geometry to be able to prove it. What I am going to tell you next I see intuitively to be true, and I hope that you also will see the same intuitively. However, if you do not see it, then please just accept that there are superb mathematicians (but sadly I am not one of them) who are able to demonstrate that what I am going to tell you next is true.

• I am told that with ANY three points it is possible to position a flat surface (say a piece of wood) so that all

three points lie on the flat surface. It does not matter how the three points are configured, there will always be a flat surface that can be positioned so that all three points are just touching that flat surface. In Geometry we say that three points in space are always “planar” i.e. they all lie in one plane. Now, if I were you, I would NOT spend an awful lot of time worrying about this concept. I do not know of a single good mathematician in the whole history of mankind who has disputed this fact.1 Many years ago I satisfied myself about the validity of this theorem by visualising a piece of wood being slipped into position under three points, and I could see how the piece of wood could always connect any three points no matter how they were positioned in relation to each other. My advice to you therefore is to try to visualise this for yourself and thus satisfy yourself that the theorem is true – or else just accept that the theorem is true just as you accept that the things that you are told about atoms and molecules and proteins and DNA and genes etc is true – but clearly you would not accept the tissue of lies about Iraq that Bush and Blair so deftly wove to deceive us all. In life you have to learn the hard way who tells the truth and who tells lies.

• With four or more points, the points may well be “planar” i.e. they may well all lie on the same plane (e.g. I

could draw 50/100/200/or ‘n’ trillion points all on one flat piece of wood, or I could draw them scattered in three dimensions so that they do not all lie on the same plane). With anything more than three points therefore we have two separate groupings i.e. Planar shapes and Non-planar shapes, and we can therefore construct the following table.

1 I am not going to go into Einstein’s Theories about the curvature of Space and Time because you could write what I know about Einstein’s theories on the back of half a postage stamp, and that is not what we are talking about here anyway. The stuff that I am discussing here is all very simple “Euclidean” Geometry (and if only it were “simple”!!!).

First Year 'A' Level Chemistry Page 2 of "Chapter 34 : The shapes of molecules and Ions"

In Euclidean Geometry PLANAR SHAPE NON-PLANAR SHAPE

One object/point

No shape

MUST be a Planar Shape

(cannot be anything else)

Two objects or points

Straight line

Three objects or points

Straight line, or a Triangle

Four objects or points

Square (amongst others)

This Chapter discusses these shapes

Five objects or points

Pentagon (amongst others)

Six objects or points

Hexagon (amongst others)

• I have gone up to only six objects, but I could have gone further had I wanted to. OK, let us now start talking

about the different shapes that you need to know in VSEPR theory for ‘A’ Level Chemistry. Molecular geometry (in a VERY simplified form) • You will have noticed that the shapes of houses are different. For example, the shapes of houses in Reading

are different from those in Worthing, and they in turn are different from those in Coronation Street; and English houses are different from Dutch houses, and they in turn are different from Mexican houses ......... and so on. Well, the same is true of the shapes of molecules and of ions – and the thing that determines the shape of a molecule or of an ion is the number of pairs of electrons possessed by the CENTRAL ATOM of the molecule or the ion.

• If two people in a group disliked each other intensely, then that would be like the repulsion which pairs of

electrons exert upon each other (because species that possess a similar electrostatic charge repel each other!). The two people who hated each other would sit as far apart as they could in a room and they would sit one in each of the furthest opposite corners of the room; and, if there were three people who all hated each other, then one of them could sit in the middle of one side of the room with the other two sitting one on each of the adjoining walls. [Do play about with this on a piece of paper and you will see how it could work.]

• In exactly the same way, pairs of electrons that repulse each other will push each other away so as to

maximise the permissible space that each one can occupy – and this will determine the shapes of the molecules of which they are a part. You do however need to know that the repulsion between two Lone Pairs of electrons (LP vs LP) is stronger than the repulsion between a Lone pair and a Bonded Pair of electrons (LP vs BP) and that, in turn, is stronger than the repulsion between two Bonded Pairs of electrons (BP vs BP).

LONE PAIRS EXERT THE STRONGEST FORCE OF REPULSION ON EACH OTHER!

LP vs LP > LP vs BP > BP vs BP • When I was learning ‘A’ Level Chemistry, I could never remember which was more powerful, a lone pair of

electrons or a bonded pair of electrons; and the following analogy may help to explain what is happening. A bonded pair of electrons is not as powerful as a lone pair of electrons because it is constrained by the force of attraction of the two bonding nuclei, whereas a lone pair of electrons is as free as a bird and it is therefore much stronger than a bonded pair!

First Year 'A' Level Chemistry Page 3 of "Chapter 34 : The shapes of molecules and Ions"

• There are two procedures that will allow you to answer most of the questions that you are likely to be asked about VSEPR, and they are as follows. When you have drawn the dot-and-cross diagram for the molecule or ion, then

1) First identify the CENTRAL ATOM in the molecule. 2) Then work out how many LONE pairs (LPs) and how many BONDED pairs (BPs) there are on the central

atom of the molecule. [NB For the purposes of determining the shapes of molecules and ions (and for no other purpose), double bonds and triple bonds are regarded as single bonds2; therefore in the example of Carbon Dioxide on page 4, although there are FOUR pairs of bonding electrons in the molecule, each double bond is treated as a single bond, therefore for the purposes of this exercise, there are deemed to be only TWO bonding pairs of electrons in Carbon dioxide!

3) Then from the TOTAL number of pairs of electrons (i.e. the number of LPs + the number of BPs) and from

the number of bonded pairs alone, use the table that follows to see the bond shape of the molecule and also the angle between the bonds on the central atom of the molecule. [Please note how everything is based around the central atom of the molecule!]

• We should start by noting that the BOND shape (i.e. the shape possessed by the bonded part of the

molecule) alters according to the number of bonded pairs on the central atom of the molecule or ion. [NB If there are just two atoms bonded together, then the overall or orbital shape would be linear but there would be no bond shape because there would be no central atom. That fact really is rather important!]

TOTAL number of No of bonded No of Example Bond Shape of Bond pairs of electrons pairs lone pairs the molecule3 Angle in the outermost shell (approx.) of the central atom A) 2 2 0 CO2/HCN Linear 180˚ B1) 3 3 0 BF3/SO3/CO32– Trigonal Planar4 120˚ B2) 3 2 1 SO2/NO2–/O3 V-shaped5 < 120˚ C1) 4 4 0 CH4/NH4+/MgCl42– Tetrahedral6 109.5˚ C2) 4 3 1 NH3/H3O+/PH3 Trigonal pyramid7 107˚ C3) 4 2 2 H2O/NH2–/SF2 V-shaped 104.5˚ • The table above has omitted the overall or Orbital shape of the molecules/ions and it considers only the

bonded shape – and please note how systematically the table exhausts each combination of the number of TOTAL pairs (the Steric Number) and Bonded Pairs viz.

A) 2/2, then B) 3/3 then 3/2, and then C) 4/4 then 4/3 then 4/2, and then and if the table had been continued, the series would have gone D) 5/5 then 5/4 then 5/3 then 5/2, ........... and so on.8 • Ions have shapes (just like molecules). The above are the most common shapes of molecules that occur. I

would advise you to learn this table off BY HEART – and especially the bond angles that are associated with any particular shape. (That will save you valuable time in the exam.)

2 The sigma-bond network determines the shape of the molecule. Therefore any electrons in pi-bonds are not counted when considering the shape of the molecule. 3 The “Bond” shape is NOT the same as the overall or Orbital shape of the species. When you study the subject of shapes in greater depth, then you will see that the TOTAL number of electron pairs (i.e. the Steric Number) determines the OVERALL shape of the molecule. The number of Bonding pairs determines the bond shape. 4 A triangle that is divided into three equal zones. 5 It is also called “bent-linear”. 6 “Tetrahedron” is the Greek word for a four sided object. If you draw a pyramid with all the sides being triangles, then you will get a tetrahedron (as you will see from pages 5 and 6). 7 Please have a look at the diagrams on page 6, and you will see the difference between a “Trigonal Pyramid” and a “Tetrahedron”. 8 The overall or orbital shape for molecules or ions with a TOTAL number of 5 and 6 electron pairs on the central atom would have been a “trigonal bipyramid” and an “octahedron”.

First Year 'A' Level Chemistry Page 4 of "Chapter 34 : The shapes of molecules and Ions"

• The beauty of learning the table off by heart is that once you know the TOTAL number of electron pairs and the number of bonded pairs on the central atom, then automatically you will know the shape of the molecule or ion, and its bond angle – therefore for a very small amount of memory work, you can crack this subject wide open! Bond shapes then become an absolute doddle (and I really do mean that)!

• OK, let’s start unravelling the table on page 3 by noting some of the interesting points in it, but before I do

so, could I urge you not to panic and ‘jump the gun’ in the exam, e.g. “ BF3” and “PF3” may look like each other, but they do NOT have the same shape9. Count your Lone Pairs and your Bonded Pairs. Then you will know the shape pf the species that you are considering.

NB Everything in the table has been grouped according to the TOTAL number of electron pairs that there are

on the central atom in the molecule i.e. 1/2/3/4/etc, and this gives the OVERALL or ORBITAL shape of the molecule or ion. [In your first year of Chemistry at University, if you use this little table and develop it further, then it will be able to tell you the overall shape/the bond shape/and the bond angle of any molecule or ion – and I have developed the table for you on page 8.]

• Now let’s look at some of the shapes of the molecules that are in the table. A) LINEAR BOND SHAPE Total no. of SETS of electron pairs on the central atom of the molecule = 2. Number of Bonded pairs = 2 / Bond Shape is Linear / Bond Angle is 180˚ [e.g. CO2]. • There are double bonds between the C atom and each of the two O atoms (and each of these bonds contains

two bonding electrons), but because double bonds and triple bonds are for VSEPR purposes treated as single bonds, there are then deemed to be only TWO SETS of bonded pairs of electrons on the C atom. [NB the lone pairs are NOT on the central atom!]

• CO2 is one example and HCN/HCl/H2/Cl2/N2 are other examples of this linear bond shape, but molecules

such as H2/Cl2/N2/and ions such as CN– cannot have a bond angle because there is only one bond in the species. CN– = [ :C≡N: ]– .

Total no. of sets of electron pairs on the central atom of the molecule = 2 Number of Bonded pairs = 2 / Bond Shape is Linear / Bond Angle is 180˚ [e.g. HCN]. • There is a triple bond between the C atom and the N atom (and each bond is made up of two bonding

electrons), but because double bonds and triple bonds are here treated as single bonds, there are deemed to be only TWO SETS of bonded pairs of electrons on the C atom. [NB the lone/the unbonded pair of electrons is on N and is NOT on the central atom C, therefore it does not affect the bond shape.]

9 I will give you a hint: the total number of pairs of electrons in PF3 is 4 and one of them is a lone pair . NOW can you see how important it is to memorise the table on page 3. You do not want to waste important time trying to remember things in the exams.

C O O XO

XO

XO

XO C O O

xx

xx

xx

xx

180˚

H C N xo xo

xo x o oo

N C H

180˚

:

N is not the central atom, therefore ignore the lone pair on N.

Hydrogen Cyanide (or Prussic Acid), is a highly toxic gas.

First Year 'A' Level Chemistry Page 5 of "Chapter 34 : The shapes of molecules and Ions"

B) THE TWO BOND SHAPES IN TRIGONAL ORBITAL SHAPES (“Orbital” shape being the overall shape of the species, and not the bond shape.) B1) Total no. of SETS of electron pairs on the central atom of the molecule = 3. Number of Bonded pairs = 3 / Bond Shape is Trigonal Planar / Bond Angle is 120˚. [e.g. Boron Trifluoride, BF3, SO3, CO32–, NO3– (the salt of the acid HNO3), and also C2H4, Ethene, where

each C atom is (in turn) regarded as a central atom having three satellite species bonded to it.] • The B atom is at the centre of the molecule, and there are three sets of bonded pairs and no lone pairs on the

central atom of the molecule. The three angles at the centre are all equal and each one equals 120˚ (because 360˚ ÷ 3 = 120˚, and 360˚ is the measure of a circle). NB The lone pairs on F are NOT on the central B atom, but they are instead on the peripheral F atoms – therefore (because of the symmetry between each of the F atoms) the Lone Pairs do not affect the bond angle!

B2) Total no. of SETS of electron pairs on the central atom of the molecule = 3 Number of Bonded pairs = 2 / Bond Shape is “V-shaped” / Bond Angle is <120˚. [e.g. Sulphur Dioxide, SO2, or the Nitrite ion NO2–]. • SO2 is like BF3 in that the overall shape of the molecule is trigonal planar, but the BONDED shape (i.e. the

shape described by the bonds) is V-shaped. There are double bonds between the S atom and each of the two O atoms, but because double bonds and triple bonds are here treated as single bonds, there are deemed to be only TWO SETS of bonded pairs of electrons on the S atom. However, the lone pair of electrons on the S atom repels the two bonded pairs and thus depresses the bond angle (making it less than 120˚!) [NB Here the lone pair IS on the central atom, therefore it DOES affect the bond angle!]

• Strictly speaking, SO2 could/should be represented as in the diagram below, but I have as yet not told you

anything about Resonance, so do not worry about it.

You might want to watch Dr Wayne Breslyn’s short (but excellent) video on SO2 at https://www.youtube.com/watch?v=REugD0mJxPk .

B

F

F F

xo

xo xo

xxxx

xx

xx

xx xx

xxxx

xx

B

F

F F

120˚

120˚

120˚

S

O

xx

Lone Pair <–––

S

O

xx oo xx oo xx

xx xx

xx

oo

O O

< 120˚

Here, Sulphur has an expanded octet. The bond angle is less than 120˚ because the lone pair is repelling the bonded pairs.

First Year 'A' Level Chemistry Page 6 of "Chapter 34 : The shapes of molecules and Ions"

C) DIFFERENT BOND SHAPES IN TETRAHEDRAL ORBITAL SHAPES C1) Total no. of sets of electron pairs on the central atom of the molecule = 4 Number of Bonded pairs = 4 / Bond Shape is Tetrahedral / Bond Angle is 109.5˚. [e.g. Methane, CH4 / NH4+ / SO42– / BF4– (BF3 + F–) ––> BF4– (BF3 is an example of a Halogen carrier)] • The central atom in this molecule is the C atom, and it is located in the burial chamber at the heart of a

symmetrical pyramid10 that, including its base, has FOUR faces (and its shape is that of a “Tetrahedron”)11 and there is then one H atom at each of the four corners of the pyramid. [NB There are NO lone pairs anywhere in this molecule.]

NB There are SIX 109.5˚ bond angles in this molecule. If you are doing Probability in ‘A’ Level Maths, then it

would be a nice little exercise for you to verify this statement using Probability Theory (there are four H atoms and you are choosing two at a time to) measure the bond angle – otherwise, please verify the statement by counting all the bonds involved (on your fingers if necessary!).

C2) Total no. of sets of electron pairs on the central atom of the molecule = 4 Number of Bonded pairs = 3 / Bond Shape is a Trigonal Pyramid / Bond Angle is 107˚. [e.g. Ammonia, NH3 or H3O+ (H2O + H+ ––> H3O+). In Europe we tend to write this as H+ (aq).]

• Please notice the difference between CH4 and NH3! The central atom in NH3 is the N atom, and it is located in the burial chamber at the heart of the pyramid, and there is then one H atom at each of the three bottom corners of the pyramid – and the top corner of the pyramid is where the lone pair of electrons is located! CH4 therefore has the shape of a Tetrahedron whereas NH3 has the overall shape of a Tetrahedron but has the BONDED shape of a Trigonal Pyramid! [NB The lone pair of electrons IS on the central atom, and it therefore affects the bond angle – it pushes the angle down below 109.5˚ to 107˚.]12

• Please verify the statement about the number of bond angles (either by use of Probability Theory, or by

counting). Each bond angle here is 107˚. • All these molecules are important, but the next one (Water) is a very important molecule. (It is one of the

most abundant molecules on the face of the planet!)

10 I used to think of it as the place where an Egyptian Pharaoh would have been buried in his/her pyramid. 11 The noun is “tetrahedron” and the adjective is “tetrahedral”. 12 All these bond angles are approximations, and you are not expected to know the exact bond angle.

C

H H

H H

xo

xo

xo

xo C

H

H H

H

H

H

H

H

C 109.5˚ 109.5˚

109.5˚ 109.5˚

The "bond angle" is the angle between each H – C – H bond. Every bond angle here is 109.5˚.

H

H

xx

H

N H H

H

xo

xo

xo

xx

The "bond angle" is the angle between each H – N – H bond. Each of these THREE bond angles is 107˚.

N

107˚ 107˚

N

H

H

H

..

First Year 'A' Level Chemistry Page 7 of "Chapter 34 : The shapes of molecules and Ions"

C3) Total no. of sets of electron pairs on the central atom of the molecule = 4 Number of Bonded pairs = 2 / the overall shape is tetrahedral, but the BOND Shape is V-shaped,

and the bond Angle is 104.5˚ (e.g. Water, H2O : this being the shape possessed by a WATER molecule). • The central O atom has 6 electrons in its outermost shell, while the two H atoms have 1 electron each in their

outermost shells. This gives 2 bonded pairs of electrons and 2 lone pairs of electrons on the Oxygen atom. • Since there are 4 pairs of electrons, the overall or orbital shape of the molecule is that of a Tetrahedron, but

since there are only 2 bonded pairs of electrons, the bonded shape of the molecule is that of a “V” because there are now TWO Lone Pairs repelling the two Bonded Pairs. (In NH3 there is only ONE Lone Pair.) I’ll say that again : The orbital or overall shape of a water molecule is tetrahedral, but the bonded shape is that of a “V”. [Some scientists call this a “bent linear” shape.]

• The O atom sits right inside at the heart of the pyramid where the Egyptian monarchs were buried. [NB There is only one bond angle in H2O.] • When I was learning Chemistry, I never really understood the shape of a water molecule until I got some

cardboard (I was born in 1940 so I missed out on Blue Peter!) and made a model of this molecule. It was only then that I realised that two of the corners of the pyramid are where the two lone pairs of electrons are located, and that if I put the model on a table with these two corners in the air, then the two corners which will be in contact with surface of the table are where the H atoms are located! You might want to get an empty cardboard carton from Tesco/Sainsburys/wherever and try making the model for yourself. You could also try typing “bond angles CH4” into your search engine and then click on “images for bond angles CH4”.

Restatement of the procedure for ascertaining the shape of a molecule or ion • The procedure for working out the shape of a molecule or ion is 1) Draw the dot-and-cross diagram (or better still do the Lewis diagram, cf. Blog dated 20th October 2019) for

the species. 2) Work out how many lone pairs and how many bonded pairs there are on the central atom of the

molecule and for this purpose, double bonds and triple bonds MUST BE regarded as single bonds! 3) Work out the bonded shape of the molecule from the number of electron pairs and bonded pairs on

the central atom of the molecule. NB Please remember that the overall shape of a molecule or ion will be determined by the TOTAL number of

pairs of electrons and the Bonded shape will then be determined by the number of Bonded pairs. The diagrams on page 9 and 10 will clarify the table on page 8.

H

xx

xx

H

xo xo

H H

O

xx xx

A water molecule has a THREE dimensional structure – a fact not conveyed by this diagram.

O

H H ..

.. O

The "bond angle" is the angle in the H – O – H bond.The bond angle is 104.5˚.

104.5˚

In theory, this diagramshould show the 3Dnature of the species.

First Year 'A' Level Chemistry Page 8 of "Chapter 34 : The shapes of molecules and Ions"

• As you can see, “The shapes of molecules/ions” is a complicated topic, but once you have learnt the table on page 3 off by heart, then it becomes child’s play (and it becomes a matter of memory rather than one of intelligence)! It’s a “no-brainer”. Once you have memorised the table on page 3, you don’t need a brain after that. I have set out below the table that some of you might want to have if you are going on to read Chemistry for your degree.

• NB From a VSEPR point of view, unless an atom has bonded to another atom, it cannot have a shape.

All that it will have is lone pairs. - Strictly speaking, VSEPR theory requires there to be a central atom before there is a VSEPR shape. - An Hydrogen atom has a spherical shape (a nucleus ‘surrounded’ by an encircling electron) – but from

the point of view of VSEPR theory it has no VSEPR shape. - Two H atoms (H–H) bonded together would have a linear shape, but again from a VSEPR point of view,

H2 has no central atom therefore it has no VSEPR shape. TOTAL no. of Bonded Lone Example Orbital Bonded Bond Angle electron pairs Pairs Pairs Shape Shape (approximate) 1 = (0 + 1) Non-bonded atom (e.g. He) therefore it cannot have a shape. –––––––––––––––––––––––––––––––––– 2 = (2 + 0) CO2/HCN Linear Linear 180˚ 2 = (1 + 1) No central atom, no bond angle. –––––––––––––––––––––––––––––––––– 3 = (3 + 0) BF3 Planar Trigonal Planar 120˚ 3 = (2 + 1) O3/SO2 Planar V-shaped < 120˚ 3 = (1 + 2) No central atom, no bond angle. –––––––––––––––––––––––––––––––––– 4 = (4 + 0) CH4/NH4+ Tetrahedron13 Tetrahedron 109.5˚ 4 = (3 + 1) NH3/H3O+ Tetrahedron Trigonal Pyramid 106.7˚ 4 = (2 + 2) H2O/SF2 Tetrahedron V-shaped 104.5˚ 4 = (1 + 3) No central atom, no bond angle. ––––––––––––––––––––––––––––– 5 = (5 + 0) PCl5 All of Trigonal Bipyramid 120˚/90˚ 5 = (4 + 1) SF4 these are Seesaw 120˚/90˚ 5 = (3 + 2) ClF3 Trigonal Deformed ‘T’ Differing 5 = (2 + 3) XeF2/I3– Bipyramids14 Linear 180˚ 5 = (1 + 4) No central atom, no bond angle. –––––––––––––––––––––––––––––––––– 6 = (6 + 0) SF6 Octahedron Octahedral 6 = (5 + 1) BrF5/IF5 Octahedron Square pyramidal 6 = (4 + 2) XeF4 15 Octahedron Square planar 6 = (3 + 3) Octahedron T-shaped 6 = (2 + 4) Octahedron Linear 6 = (1 + 5) No central atom, no bond angle. NB1 If there is just one bonded pair in a species, then there cannot be a Bond Angle, because there will be only one bond –

and there needs to be at least two bonds to form an angle! NB2 The orbital shape remains the same throughout each grouping e.g. in “TOTAL no. of pairs = 3” the orbital shape will always be “Planar”, and in “TOTAL no. of pairs = 4” the orbital shape will always be “Tetrahedron”, and in “TOTAL no. of pairs = 5” the orbital shape will always be “Trigonal Bipyramid”, and in “TOTAL no. of pairs = 6” the orbital shape will always be “Octahedron”, and so on.

13 “Tetrahedron” is the Greek word for a four sided object. If you draw a pyramid with all the sides being triangles, then you will get a tetrahedron. 14 Two trigonal pyramids that have their bases stuck to each other, and one pyramid is therefore “upside down”. 15 The ion [ICl4] – would also be a good example.

First Year 'A' Level Chemistry Page 9 of "Chapter 34 : The shapes of molecules and Ions"

• You may find the following table of some use. You will find it on

http://en.wikipedia.org/wiki/VSEPR_theory where “E” is the central atom and “X” is an atom that is bonded to it.

• There are some nice diagrams on the Department of Chemistry, University of Florida site at http://www.chem.ufl.edu/~myers/chm2045/shapes.htm

<–– e.g. SO2 has a bond angle of 119˚.

<–– H2O is a good example of this.

First Year 'A' Level Chemistry Page 10 of "Chapter 34 : The shapes of molecules and Ions"

Because the diagrams below are in colour, you may also find this very useful indeed (but do not worry too much about the hybrid spn orbitals that are shown in blue).

You could also try Michigan State University’s excellent website http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/intro3.htm I have not taught you anything about “resonance” as yet, but do not worry about that.

First Year 'A' Level Chemistry Page 11 of "Chapter 34 : The shapes of molecules and Ions"

• And still on shapes, this one is from London Metropolitan University (Dr N Chatterton)

Dipole Moments and Symmetry • I have already mentioned that it was possible for a molecule to have polarised bonds but (if the molecule

was symmetrical) then the molecule itself would not be polarised (e.g. BF3 has polarised bonds but it is still not a polarised molecule). If you are not doing ‘A’ Level Maths then you will have to ask one of your friends (or come and see me in one of your breaks) about Trigonometrical expressions which can be summarised by the memory tag SOH CAH TOA. (Just pronounce it as though it were an English word.)

• BF3 is a planar symmetrical species with bond angles of 120˚ therefore two of the B–F bonds can be reduced

to in effect one bond in the same line as the third bond by using the equations Cos =Adjacent ÷ Hypotenuse therefore Adjacent = Hypotenuse x Cos 60 A = H x 0.5, and 2 x 0.5 = 1;

and, since this virtual bond points in exactly the opposite direction to the third bond, the two now cancel each other out. It is therefore true that the bonds in BF3 are polarised, while the overall molecule is not polarised.

• VSEPR theory shows why this does not apply to NH3 because NH3 is not a symmetrical molecule. Instead it

is a tetrahedral pyramidal molecule with a lone pair of electrons at one vertex and with bond angles of 106.7˚ (which for ‘A’ Level purposes you can call 107˚). The base of the pyramid is symmetrical, but the overall molecule is not symmetrical therefore the trigonometrical dipole moments do not cancel each other out in all the spherical directions.

• It is not that difficult to apply 3-D as opposed to planar trigonometry to show that although CCl4 has

polarised bonds, nevertheless because it is symmetrical, the molecule itself is not polar. CCl4 is therefore NOT a polarised molecule, but NH3 is polarised.

• Could you therefore please accept that species that have polarised bonds but are symmetrical in three

dimensions are NOT polarised species.

First Year 'A' Level Chemistry Page 12 of "Chapter 34 : The shapes of molecules and Ions"

APPENDIX • I do not want you to forget about Lewis structures, therefore let us draw HClO (Hypochlorous Acid) and

HCN (Hydrogen Cyanide). • Before we do so, I have to alert you to the fact that all that we have learnt so far is ‘A’ Level Chemistry.

Please therefore just accept that even though Oxygen is not the least electronegative element in these molecules, nevertheless the least unstable way of drawing these molecules is to put the ‘O’ atom in the centre of the molecule.

• Now could you please go to https://www.youtube.com/watch?v=SU9ohsz4iTs to see how DrWayne Breslyn

has drawn the Lewis structure of HClO (where the ‘O’ atom goes in the middle of the molecule).

and to https://www.youtube.com/watch?v=Ap93tYTN8C4 to see how Dr Breslyn has drawn HCN (where

again, ‘C’ is the central atom).